Chapter 11. Rotational Dynamics As we did for linear (or translational) motion, we studied kinematics (motion without regard to the cause) and then dynamics (motion with regard to the cause), we now proceed in a similar fashion
We know that forces are responsible for linear motion
We will now see that rotational motion is caused by torques
Consider a wrench of length L=r
F
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τ = rF
€
r
Torque
Units of N m
Not work or energy, do not use Joules
is called the lever arm. Must always be perpendicular to the force
If the force is not perpendicular to the level arm, we need to find the component that is perpendicular (either the force or the lever arm)
€
r
φ
F F sinφ
€
τ = rF sinφ
If φ=0, then the torque is zero
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r Therefore, torques (force times length) are responsible for rotational motion
Newton’s 2nd Law for Rotational Motion
F=Ft
r m
The torque for a point-particle of mass m a distance r from the rotation axis is
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τ = rFtFt = mat, at = rα⇒ τ = matr = mr2α
Axis of rotation
Define I = mr2 = Moment of Inertia for a point particle; a scalar, units of kg m2
A rigid body is composed of many, many particles
of mass mi which are ri from the axis of rotation
Each of these masses creates a torque about the axis of rotation
....
2222
2111
ατατrmrm
== m1
m2
r1
r2
Rotation axis
Sum up all torques due to all particles
∑∑
∑ ∑
==
=
ατ
ατ
IrmIrm
ii
iii2
2 α is the same for all particles
I =moment of inertia for the rigid body. It is different for different shaped objects and for different axes of rotation. Table 10.1.
For a thin rod of mass M and length L
L/2 L 2121 MLI =
231MLI =
The last equation is Newton’s 2nd Law for rotation. Compare to the translational form
Example Problem
A rotating door is made of 4 rectangular panes each with a mass of 85 kg. A person pushes on the outer edge of one pane with a force of 68 N, directed perpendicular to the pane. Determine the door’s α.
amFI =⇔= ∑∑ ατ
L
F IFLIFL
I
=⇒=
=∑αα
ατ
Given: L = 1.2 m, mpane=85 kg, F=68 N
From Table 10.1, moment of inertia for a thin rectangular rod is (same as a thin sheet)
Since there are 4 panes.
2342
31 MLIMLI doorpane =⇒=
2srad
m kgN
234
50.050.0)m 2.1)(kg 85(4
)N 68(343
====
==
MLF
MLFL
IFL
α
α
Example Problem The parallel axis theorem provides a useful way to calculate I about an arbitrary axis. The theorem states that I = Icm + MD2, where Icm is the moment of inertia of an object (of mass M) with an axis that passes through the center of mass and is parallel to the axis of interest. D is the perpendicular distance between the two axes. Now, determine I of a solid cylinder of radius R for an axis that lies on the surface of the cylinder and perpendicular to the circular ends.
Solution: The center of mass of the cylinder is on a line defining the axis of the cylinder
R
221 MRIcm =From Table 10.1:
From the parallel axis theorem with D=R:
Apply to thin rod:
cm
22322
212 MRMRMRMDII cm =+=+=
L/2 L 2121 MLIcm =
231
22
1212
2MLLMMLMDII cm =
+=+=
Rotational Work
For translational motion, we defined the work as
For rotational motion
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W = Fddθ
r
r s = arc length
= rθ
s
€
d
€
F d
€
W = Fdd = Ftrθ = (Ftr)θWR = τ θ
Ft
Units of N m or J when θ is in radians
Rotational Kinetic Energy For translational motion, the K was defined
22212
2121
)(,
ωωω
mrrmKrmK t
==== v since v2
For a point particle I = mr2, therefore
Or for a rigid body
For a rigid body that has both translational and rotational motion, its total kinetic energy is
The total mechanical energy is then
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Krot = 12 Iω
2
€
Krot = 12 miri
2ω 2 = 12∑ Iω 2 = Krot
€
Ktotal = K + Krot = 12mv
2 + 12 Iω
2
€
Etotal = 12mv
2 + 12 Iω
2 + mgy + 12 kx
2
Example A car is moving with a speed of 27.0 m/s. Each wheel has a radius of 0.300 m and a moment of inertia of 0.850 kg m2. The car has a total mass (including the wheels) of 1.20x103 kg. Find (a) the translational K of the entire car, (b) the total Krot of the four wheels, and (c) the total K of the car.
Solution:
Given: vcar = 27.0 m/s, mcar = 1.20x103 kg, rw = 0.300 m, Iw = 0.850 kg m2
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K = 12 mcarvcar
2 = 12 (1.2x103)(27.0)2
K = 4.37x105 J
Krot = 12 Iω
2 = 12 Iw
vt
rw
2
= 12 Iw
vcar
rw
2
Krot = 12 (0.850)(27.0 /0.300)2
= 3.44x103 JKwheels = 4Krot =1.38x104 JKtotal = K + Kwheels= 4.37x105 +1.38x104 = 4.51x105 J
a)
b)
c)
Example Problem A tennis ball, starting from rest, rolls down a hill into a valley. At the top of the valley, the ball becomes airborne, leaving at an angle of 35° with respect to the horizontal. Treat the ball as a thin-walled spherical shell and determine the horizontal distance the ball travels after becoming airborne.
1.8 m θ
x
0
1 2
3
4
Solution:
Given: v0 = 0, ω0 = 0, y0 = 1.8 m = h, y1 = y2 = y4 = 0, θ2 = 35°, x2 = 0, I=(2/3)MR2
Find: x4 ?
Method: As there is no “friction” or air resistance in the problem, therefore no non-conservative forces, we can use conservation of mechanical energy
1222
1222
12
212
1212
11
0
)0(v)0(vEmgImE
mgImEmghE
=++=++=
=
ωω
mgyImEtotal ++= 2212
21 v ω
m/s 6.48)/56(9.80)(1.
5/6 v
vvv
)/v)((v
vvbut v ,v
Since ,v v
2
226
5223
1222
1
22
232
212
221
2
222
1222
1
210
2121
==
=⇒
=+=
+=
=
=+=
==
==⇒
gh
gh
rmrmmgh
rImmghEEE
t
t ωω
ωω
Velocity of ball equals tangential velocity at edge of ball
Now, use 2D kinematic equations for projectile motion
m 0.25
35sin35cos)8.1(125
sincos12g
sincos25
6g
sincos2v/sinvcos2v/v2v
v)vv(/2vv2
vv0 ),(vv
vsinvvvcosvv
4
224
222222
4
2222224
4244221
24
2442
422
22424
4222
4222
=
==
==
===++=
=⇒−=−−=−
=−−=−==
==
x
hx
ghx
ggxttxxgtgt
gttttg
yx
xxx
yy
yy
yy
yy
xx
θθ
θθθθθθ
θθ
v2
v4
Example Problem A mass m1 (=15.0 kg) and a mass m2
(=10.0 kg) are suspended by a pulley that has a radius R (=10.0 cm) and a mass M (=3.00 kg). The cord has negligible mass and causes the pulley to rotate without slipping. The pulley rotates about its axis without friction. The masses start from rest at a distance h (=3.00 m) part. Treating the pulley as a uniform disk, determine the speeds of the two masses as they pass each other.
